Running select filename from dbo.sysfiles returns short foldernames with tilde

I am running the following statement in SQL Server 2014(in Windows 2012R2):

select filename from dbo.sysfiles where fileid = 1 

However it returns:

c:\progra~1\micros~1\mssql1~1.sql\mssql\data\master~1.mdf 

I wish for the full path without the tilde.

(e.g C:\Program Files\Microsoft SQL Server\MSSQL12.SQL2014\MSSQL\DATA\master.mdf)

I tried upgrading the SQL Server 2014 Express to 2016 Express but it made no difference.

Any ideas? Thanks. Dave

I can’t bind the ` (tilde) key as hotkey for DoTA 2 scoreboard?

I’m using English (US,alt,intl) and for whatever reason the game doesn’t recognize when I use the ` key. Any other keys work fine. Of course, changing my keyboard input to another language made the button where my ` is be something else and it works but it’s annoying to switch languages. Any ideas how to use tilde in DoTA 2?

If the Hausforff dimension of the graph of a function $u$ is $N$ and $\tilde u = u$ a.e. then $\dim_H \mathrm{graph} \, \tilde u = N$ too

Let $ u:\Omega \subset \mathbb{R}^N \to \mathbb{R}^M$ be a function such that the Hausdorff dimension of its graph is $ N$ .

Let $ \tilde u = u$ a.e. Is it true that the Hausdorff dimension of the graph of $ \tilde u$ is also $ N$ ?

Sidebar and dash missing – I can’t find the tilde button

I’ve seen others with the same issue that I am having and I see a potential solution however for some reason in the process my keyboard has changed configuration and the @ Button is no longer shift+2 but shift+”

I think this now means of can’t type the tilde button which is included in the compizconfig solution.

Any help I’d be very grateful. I’ve lost my mailbox from the desktop.

It’s situation critical

14.04 is what I’m using by the way

Find a modified coupling $((X_n,\tilde Y_n))_{n∈ℕ_0}$ with the same coupling time $τ$ and $\tilde Y_n=X_n$ for $n≥τ$ in the coupling lemma

Let

  • $ (\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $ (E,\mathcal E)$ be a measurable space
  • $ (X_n)_{n\in\mathbb N_0}$ and $ (Y_n)_{n\in\mathbb N_0}$ be $ (E,\mathcal E)$ -valued time-homogeneous Markov chains on $ (\Omega,\mathcal A,\operatorname P)$ with common transition kernel $ \kappa$ and $ $ Z_n:=(X_n,Y_n)\;\;\;\text{for }n\in\mathbb N_0$ $
  • $ \mathcal F^X$ , $ \mathcal F^Y$ and $ \mathcal F^Z$ denote the filtraiton generated by $ X$ , $ Y$ and $ Z$ , respectively

It’s easy to see that $ $ \tau:=\inf\left\{n\in\mathbb N_0:X_n=Y_n\right\}$ $ is an $ \mathcal F^Z$ -stopping time and hence $ $ \tilde Y_n:=1_{\left\{\:n\:<\:\tau\:\right\}}Y_n+1_{\left\{\:n\:\ge\:\tau\:\right\}}X_n\;\;\;\text{for }n\in\mathbb N_0$ $ is $ \mathcal F^Z$ -adapted. Moreover, $ \mathcal F^Z=\mathcal F^X\vee\mathcal F^Y$ .

How can we show that $ \tilde Y$ is a time-homogeneous Markov chain with the same distribution as $ Y$ ?

I guess the basic idea is that $ Z$ is clearly a time-homogeneous Markov chain with transition kernel $ \pi$ satisfying $ $ \pi((x,y),B_1\times B_2)=\kappa(x,B_1)\kappa(y,B_2)\;\;\;\text{for all }x,y\in E\text{ and }B_i\in\mathcal E\tag1.$ $ Since $ \mathbb N_0$ is countable, $ Z$ is strongly Markovian at $ \tau$ and hence $ $ 1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname E\left[f\left(\left(Z_{\tau+n}\right)_{n\in\mathbb N_0}\right)\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:<\:\infty\:\right\}}(\pi f)(Z_\tau)\;\;\;\text{almost surely}\tag2,$ $ where $ \pi f:=\int\pi(\;\cdot\;,{\rm d}z)f(z)$ , for all bounded and $ (\mathcal E\otimes\mathcal E)^{\otimes\mathbb N_0}$ -measurable $ f:(E\times E)^{\mathbb N_0}\to\mathbb R$ . So, if $ k\in\mathbb N_0$ , $ n_0,\ldots,n_k\in\mathbb N_0$ with $ 0=n_0<\cdots<n_k$ and $ B\in\mathcal E^{\otimes k}$ , we obtain \begin{equation}\begin{split}1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(\tilde Y_{\tau+n_1},\ldots,\tilde Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\bigotimes_{i=1}^k\kappa^{n_i-n_{i-1}}(X_\tau,B)\&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(Y_{\tau+n_1},\ldots,Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]\end{split}\tag3\end{equation} almost surely.

However, it’s neither clear to me how we can conclude that $ \tilde Y$ is Markovian (with respect to its generated filtration) nor why it has the same distribution as $ Y$ .

Clearly, the distribution of $ Y$ is uniquely determined by the finite-dimensional distributions $ \operatorname P\left[\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]$ (and the same applies to $ \tilde Y$ ). Moreover, we may write $ $ \operatorname P\left[\left(\tilde Y_{n_1},\ldots,\tilde Y_{n_k}\right)\;\cdot\;\right]=\operatorname P\left[n<\tau,\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]+\operatorname P\left[n\ge\tau,\left(X_{n_1},\ldots,X_{n_k}\right)\;\cdot\;\right]\tag4.$ $ Many pieces, I’m not able to combine.

Find a modified coupling $((X_n,\tilde Y_n))_{n∈ℕ_0}$ with the same coupling time $τ$ and $\tilde Y_n=X_n$ for $n≥τ$ in the coupling lemma

Let

  • $ (\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $ (E,\mathcal E)$ be a measurable space
  • $ (X_n)_{n\in\mathbb N_0}$ and $ (Y_n)_{n\in\mathbb N_0}$ be $ (E,\mathcal E)$ -valued time-homogeneous Markov chains on $ (\Omega,\mathcal A,\operatorname P)$ with common transition kernel $ \kappa$ and $ $ Z_n:=(X_n,Y_n)\;\;\;\text{for }n\in\mathbb N_0$ $
  • $ \mathcal F^X$ , $ \mathcal F^Y$ and $ \mathcal F^Z$ denote the filtraiton generated by $ X$ , $ Y$ and $ Z$ , respectively

It’s easy to see that $ $ \tau:=\inf\left\{n\in\mathbb N_0:X_n=Y_n\right\}$ $ is an $ \mathcal F^Z$ -stopping time and hence $ $ \tilde Y_n:=1_{\left\{\:n\:<\:\tau\:\right\}}Y_n+1_{\left\{\:n\:\ge\:\tau\:\right\}}X_n\;\;\;\text{for }n\in\mathbb N_0$ $ is $ \mathcal F^Z$ -adapted. Moreover, $ \mathcal F^Z=\mathcal F^X\vee\mathcal F^Y$ .

How can we show that $ \tilde Y$ is a time-homogeneous Markov chain with the same distribution as $ Y$ ?

I guess the basic idea is that $ Z$ is clearly a time-homogeneous Markov chain with transition kernel $ \pi$ satisfying $ $ \pi((x,y),B_1\times B_2)=\kappa(x,B_1)\kappa(y,B_2)\;\;\;\text{for all }x,y\in E\text{ and }B_i\in\mathcal E\tag1.$ $ Since $ \mathbb N_0$ is countable, $ Z$ is strongly Markovian at $ \tau$ and hence $ $ 1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname E\left[f\left(\left(Z_{\tau+n}\right)_{n\in\mathbb N_0}\right)\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:<\:\infty\:\right\}}(\pi f)(Z_\tau)\;\;\;\text{almost surely}\tag2,$ $ where $ \pi f:=\int\pi(\;\cdot\;,{\rm d}z)f(z)$ , for all bounded and $ (\mathcal E\otimes\mathcal E)^{\otimes\mathbb N_0}$ -measurable $ f:(E\times E)^{\mathbb N_0}\to\mathbb R$ . So, if $ k\in\mathbb N_0$ , $ n_0,\ldots,n_k\in\mathbb N_0$ with $ 0=n_0<\cdots<n_k$ and $ B\in\mathcal E^{\otimes k}$ , we obtain \begin{equation}\begin{split}1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(\tilde Y_{\tau+n_1},\ldots,\tilde Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\bigotimes_{i=1}^k\kappa^{n_i-n_{i-1}}(X_\tau,B)\&=1_{\left\{\:\tau\:<\:\infty\:\right\}}\operatorname P\left[\left(Y_{\tau+n_1},\ldots,Y_{\tau+n_k}\right)\in B\mid\mathcal F_\tau\right]\end{split}\tag3\end{equation} almost surely.

However, it’s neither clear to me how we can conclude that $ \tilde Y$ is Markovian (with respect to its generated filtration) nor why it has the same distribution as $ Y$ .

Clearly, the distribution of $ Y$ is uniquely determined by the finite-dimensional distributions $ \operatorname P\left[\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]$ (and the same applies to $ \tilde Y$ ). Moreover, we may write $ $ \operatorname P\left[\left(\tilde Y_{n_1},\ldots,\tilde Y_{n_k}\right)\;\cdot\;\right]=\operatorname P\left[n<\tau,\left(Y_{n_1},\ldots,Y_{n_k}\right)\;\cdot\;\right]+\operatorname P\left[n\ge\tau,\left(X_{n_1},\ldots,X_{n_k}\right)\;\cdot\;\right]\tag4.$ $ Many pieces, I’m not able to combine.